A) \[11\]
B) \[12\]
C) \[10\]
D) \[14\]
Correct Answer: B
Solution :
Let \[y=4{{e}^{2x}}+9{{e}^{-2x}}\] \[\therefore \] \[\frac{dy}{dx}=8{{e}^{2x}}-18{{e}^{-2x}}\] \[\Rightarrow \] x\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=16{{e}^{2x}}+36{{e}^{-2x}}\] For minimum, put \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[8{{e}^{2x}}-18{{e}^{-2x}}=0\] \[\Rightarrow \] \[8{{e}^{2x}}=18{{e}^{-2x}}\] \[\Rightarrow \] \[{{e}^{4x}}=\frac{18}{8}=\frac{9}{4}\] \[\Rightarrow \] \[4x=\log \frac{9}{4}\] \[\Rightarrow \] \[x=\frac{1}{4}\log \frac{9}{4}\] At \[x=\frac{1}{4}\log \frac{9}{4}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}>0\] \[\therefore \] Minimum value at \[x=\frac{1}{4}\log \frac{9}{4}\] is \[y=4{{e}^{2\left( \frac{1}{4}\log \frac{9}{4} \right)}}+9{{e}^{-2\left( \frac{1}{4}\log \frac{9}{4} \right)}}\] \[=4{{e}^{\log {{\left( \frac{9}{4} \right)}^{1/2}}}}+9{{e}^{\log {{\left( \frac{9}{4} \right)}^{-1/2}}}}\] \[=4{{\left( \frac{9}{4} \right)}^{1/2}}+9{{\left( \frac{9}{4} \right)}^{-1/2}}\] \[=4.\frac{3}{2}+9.\frac{2}{3}=6+6=12\]
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